Difference between revisions of "2013 AMC 8 Problems/Problem 8"
Larryflora (talk | contribs) (→Solution 1) |
Larryflora (talk | contribs) |
||
| Line 7: | Line 7: | ||
https://youtu.be/6xNkyDgIhEE?t=44 | https://youtu.be/6xNkyDgIhEE?t=44 | ||
| − | ==Solution | + | ==Solution 1== |
First, there are <math>2^3 = 8</math> ways to flip the coins, in order. | First, there are <math>2^3 = 8</math> ways to flip the coins, in order. | ||
| Line 16: | Line 16: | ||
Therefore, the probability of flipping at least two consecutive heads is <math>\boxed{\textbf{(C)}\ \frac38}</math>. | Therefore, the probability of flipping at least two consecutive heads is <math>\boxed{\textbf{(C)}\ \frac38}</math>. | ||
| − | ==Solution | + | ==Solution 2== |
Let's figure it out by complementary counting. | Let's figure it out by complementary counting. | ||
Revision as of 08:24, 3 August 2021
Problem
A fair coin is tossed 3 times. What is the probability of at least two consecutive heads?
Video Solution
https://youtu.be/6xNkyDgIhEE?t=44
Solution 1
First, there are 
ways to flip the coins, in order.
The ways to get no one head are HTH and THH.
The way to get three consecutive heads is HHH.
Therefore, the probability of flipping at least two consecutive heads is 
.
Solution 2
Let's figure it out by complementary counting.
First, there are ways to flip the coins, in order.
Secondly, what we don't want are the ways without getting two consecutive heads: TTT, HTH, and THT. Then we can find out the probability of these three ways of flipping is 
, 
,and , respectively. So the left is exactly the probability of flipping at least two consecutive heads: 
. It is the answer . ----LarryFlora
See Also
| 2013 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 7 |
Followed by Problem 9 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
| All AJHSME/AMC 8 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
